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Q: How many 4-digit numbers are possible if the hundreds digit is 8 and if repetition of digits is allowed?

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There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.

290

If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210

if the repetition is allowed the there is 6*6*6 possible ways = 216

Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.

64 if repetition is allowed.24 if repetition is not allowed.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

When the repetition is allowed: 4*4*4 = 64 numbers. Without repetition: 4*3*2 = 24 numbers.

There are 504 of them.

If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.

Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77

There are 5*5*5 = 125 such numbers.

-123456787

20 of them, if repetition is not allowed.

9,999 if zeroes can be used than it would be 10,999

There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.

If repetition is allowed [such as 122 or 555], then there are 5Â³ = 125 possibilities.

Three numbers can be arranged in 27 different sequences if repetition is allowed, and in 6 different sequences if it's not.

10^7 if the repetition of digits is allowed. 9*8*7*6*5*4*3 , if the repetition of digits is not allowed.

24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.

Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.

Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2. If repetition is allowed, 1212

If order is important and repetition is allowed, then : 000 through 999 (1000 possible). If no leading zeros, then start at 100 through 999 (900 possible). See related link, if you need to limit to no repetition, or order doesn't matter (like if 123 is the same as 231).

125

It is 415968.